= Trapped-ion fluorescence count model
{title2=$P(K=k\mid s)=e^{-\lambda_s}\lambda_s^k/k!$}
For an original illustrative readout model, take detected <photon> count $K$ in a fixed exposure to have a <Poisson distribution>, with stipulated means $\lambda_B=12$ for a bright state and $\lambda_D=0.2$ for a dark state. Classify $K\ge3$ as bright. Direct summation gives
$$
P(\text{dark label}\mid B)=e^{-12}\left(1+12+\frac{12^2}{2}\right)\simeq0.0005223,
$$
$$
P(\text{bright label}\mid D)=1-e^{-0.2}\left(1+0.2+\frac{0.2^2}{2}\right)\simeq0.0011485.
$$
With equally likely preparations, the average misclassification <probability> is $0.0008354$. These numbers describe only the chosen count model. Actual <fluorescence> includes short-time photon correlations, detector effects and possible state changes; it need not follow a fixed-rate <Poisson process>. Longer exposure can improve separation of count distributions while increasing unwanted <optical pumping>, metastable decay or recoil.
If the prepared bright population is $p$, let $\alpha$ and $\beta$ denote the two conditional errors above. The observed bright-label <probability> is $q=(1-\alpha)p+\beta(1-p)$, hence $p=(q-\beta)/(1-\alpha-\beta)$ when the denominator is nonzero. Known-state controls must estimate $\alpha,\beta$ in the actual apparatus; the formula cannot correct unknown drift. For $M$ independent preparations, the raw bright fraction has <variance> $q(1-q)/M$. Counting uncertainty, preparation error, pulse error and state-detection bias therefore enter different parts of the inference. Repeating a biased measurement reduces its sampling uncertainty without removing its bias.
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